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3 years ago

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A line segment beginning at point A(−7,−10) and ending at point B(5,8) is drawn on a coordinate plane. Point C(−3,−4) is plotted

on AB⎯⎯⎯⎯⎯⎯⎯ . Point C divides AB⎯⎯⎯⎯⎯⎯⎯ into a ratio of : . (Express the ratio in simplest integer form.)

1 answer:

riadik2000 [5.3K]3 years ago

5 0

The total x-distance from A to B is 5 - (-7) = 12.

The x-distance from C to A is -3 -(-7) = 4.

Then the x-distance from C to B is 12 - 4 = 8.

The ratio AC : CB is 4 : 8. These ratio values have a common factor of 4, so we can divide by that to get the simplest integer form:

... AC : CB = 1 : 2 . . . . . the ratio into which point C divides the line segment.

_____

If you perform the same math on y-coordinates, you should get the same answer. If you don't, it means that point C is not on the line AB.

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Determine whether each expression below is always, sometimes, or never equivalent to sin x when 0° < x < 90° ? Can someone

Lubov Fominskaja [6]

Answer:

$(a)\ \cos(180 - x)$ --- Never true

$(b)\ \cos(90 -x)$ --- Always true

$(c)\ \cos(x)$ ---- Sometimes true

$(d)\ \cos(2x)$ ---- Sometimes true

Step-by-step explanation:

Given

$\sin(x )$

Required

Determine if the following expression is always, sometimes of never true

$(a)\ \cos(180 - x)$

Expand using cosine rule

$\cos(180 - x) = \cos(180)\cos(x) + \sin(180)\sin(x)$

$\cos(180) = -1\ \ \sin(180) =0$

So, we have:

$\cos(180 - x) = -1*\cos(x) + 0*\sin(x)$

$\cos(180 - x) = -\cos(x) + 0$

$\cos(180 - x) = -\cos(x)$

$-\cos(x) \ne \sin(x)$

Hence: (a) is never true

$(b)\ \cos(90 -x)$

Expand using cosine rule

$\cos(90 -x) = \cos(90)\cos(x) + \sin(90)\sin(x)$

$\cos(90) = 0\ \ \sin(90) =1$

So, we have:

$\cos(90 -x) = 0*\cos(x) + 1*\sin(x)$

$\cos(90 -x) = 0+ \sin(x)$

$\cos(90 -x) = \sin(x)$

Hence: (b) is always true

$(c)\ \cos(x)$

If

$\sin(x) = \cos(x)$

Then:

$x + x = 90$

$2x = 90$

Divide both sides by 2

$x = 45$

(c) is only true for $x = 45$

Hence: (c) is sometimes true

$(d)\ \cos(2x)$

If

$\sin(x) = \cos(2x)$

Then:

$x + 2x = 90$

$3x = 90$

Divide both sides by 2

$x = 30$

(d) is only true for $x = 30$

Hence: (d) is sometimes true

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3 years ago

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OLEGan [10]

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

$y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{2}{9}} x-7\qquad \impliedby \qquad \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

so then we can say that

$\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{-\cfrac{2}{9}} ~\hfill \stackrel{reciprocal}{-\cfrac{9}{2}} ~\hfill \stackrel{negative~reciprocal}{-\left( -\cfrac{9}{2} \right)\implies \cfrac{9}{2}}}$

so we're really looking for the equation of a line whos slope is 9/2 and passes through (2 , 3)

$(\stackrel{x_1}{2}~,~\stackrel{y_1}{3})\qquad \qquad \stackrel{slope}{m}\implies \cfrac{9}{2} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{3}=\stackrel{m}{\cfrac{9}{2}}(x-\stackrel{x_1}{2}) \\\\\\ y-3=\cfrac{9}{2}x-9\implies \boxed{y=\cfrac{9}{2}x-6}$

now as far as the parallel line

keeping in mind that parallel lines have exactly the same slope, so we're really looking for the equation of a line whose slope is -2/9 and passes through (2 , 3)

$(\stackrel{x_1}{2}~,~\stackrel{y_1}{3})\qquad \qquad \stackrel{slope}{m}\implies -\cfrac{2}{9} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{3}=\stackrel{m}{-\cfrac{2}{9}}(x-\stackrel{x_1}{2}) \\\\\\ y-3=-\cfrac{2}{9}x+\cfrac{4}{9}\implies y=-\cfrac{2}{9}x+\cfrac{4}{9}+3\implies \boxed{y=-\cfrac{2}{9}x+\cfrac{31}{9}}$

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The origin is (0,0)

slope = (y2 - y1) / (x2 - x1)
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(0,0)...x2 = 0 and y2 = 0
now we sub
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